JEE Main 2026 (Online) 28th January Morning ShiftQuestion 1
Let $PQR$ be a triangle such that $\overrightarrow{PQ}=-2\hat i-\hat j+2\hat k$ and $\overrightarrow{PR}=a\hat i+b\hat j-4\hat k$, $a,b\in\mathbf{Z}$. Let $S$ be the point on $QR$ which is equidistant from the lines $PQ$ and $PR$. If $|PR|=9$ and $\overrightarrow{PS}=\hat i-7\hat j+2\hat k$, then the value of $3a-4b$ is $\_\_\_\_\_\_$ .
Vector Algebra
Answer$37$
Solution
$|PQ|=3$. $S$ lies on the bisector, so $\overrightarrow{PS}\parallel\frac{\overrightarrow{PQ}}3+\frac{\overrightarrow{PR}}9\parallel 3\overrightarrow{PQ}+\overrightarrow{PR}=(a-6,b-3,2)$. Matching $z$-components with $(1,-7,2)$ gives $a=7,b=-4$ (and $|PR|=9$ holds). $3a-4b=21+16=37$.